Question 1

There is a cylinder circumscribing the hemisphere such that their bases are common. Find the ratio of their volumes

Accepted Answer

R =OA = OB = Radius of hemisphere. = Radius of cylinder ABCD. Height of the cylinder = BC = radius Volume of cylinder/Volume of hemisphere = πr 3 2 3 πr 3

Question 2

The capacity of a tank of dimensions (8m x 6m x 2.5m)is

Accepted Answer

capacity of a tank = volume of tank = [ (8 x 100) x (6 x 100) x (2.5 x 100) ]/ 1000 litres = 120000 litres

Question 3

How many iron rods, each of length 7 mts and diameter 2 cms can be made out of 0.88 cubic metre of iron ?

Accepted Answer

Given length of iron rod =7 mts and and diameter of rod = 2 cms => r = 1/100 mts Therefore, Volume of one rod = πr 2 h cu. m = 22 7 × 1 100 2 × 7 cu. m = 11 5000 cu. m Given volume of iron = 0.88 cu. m Therefore, Number of rods = 0 . 88 × 5000 11 = 400.

Question 4

The areas of three consecutive faces of a cuboid are 12 cm², 20cm² and 15 cm², then the volume (in cm³) of the cuboid is

Accepted Answer

From the given data, let the length, breadth and height of the cuboid are m, n, r m x n = 12 n x r = 20 r x m = 15 Hence, m x n x n x r x r x m = 12 x 20 x 15 mnr = sqrt of (12x20x15) = 60 cub.cm.

Question 5

A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1cm when a man gets on it. The mass of the man is:

Accepted Answer

60 kg

Question 6

Water flows into a tank 200 m x 160 m through a rectangular pipe of 1.5m x 1.25 m @ 20 kmph . In what time (in minutes) will the water rise by 2 meters?

Accepted Answer

Volume required in the tank = (200 x 150 x 2) cu.m = 60000 cu.m Length of water column flown in1 min =(20 x 1000)/60 m =1000/3 m Volume flown per minute = 1.5 x 1.25 x (1000/3) cu.m= 625 cu.m. Required time = (60000/625)min = 96min

Question 7

A cylindrical rod of iron, whose height is equal to its radius, is melted and cast into spherical balls whose radius is half the radius of the rod. Find the number of balls.

Accepted Answer

Volume of rod = πr 2 h = πr 2 × r = πr 3 Volume of one spherical ball =r/2 Volume of the spherical ball = 4 3 πr 3 8 Number of balls=Volumeof rod/volumeof one ball

Question 8

A rectangular box measures internally 1.6m long, 1m broad and 60cm deep. The number of cubical box each of edge 20cm that can be packed inside the box is:

Accepted Answer

Number of blocks = 160 × 100 × 60 20 × 20 × 20 =120

Question 9

The height of the wall is 6 times its width and lenght of the wall is 7 times its height .if the volume of the wall be 16128 cu.m.its width is

Accepted Answer

Let width = x Then, height = 6x and length = 42x 42x * 6x * x = 16128 x = 4

Question 10

Which measurement unit represents volume?

Accepted Answer

All of the above

Question 11

If the radius of base and height of a cone are increased by 10%, find the percentage of increase of its volume

Accepted Answer

v1= 1 3 πr 1 2 h 1 v2= 1 3 πr 2 2 h 2 % Increase in volume = (V2-V1) / (V2*100)

Question 12

A single pipe of diameter x has to be replaced by six pipes of diameters 12 cm each. The pipes are used to covey some liquid in a laboratory. If the speed/flow of the liquid is maintained the same then the value of x is ?

Accepted Answer

Volume of water flowing through 1 pipe of diameter x = Volume discharged by 6 pipes of diameter 12 cms. As speed is same, area of cross sections should be same. Area of bigger pipe of diameter x = Total area of 6 smaller pipes of diameter 12 i.e πR 2 = 6 πR 1 2 Here R = R and R1 = 12/2 = 6 ⇒ R 2 = 6 × 6 × 6 R = 14.696 => D = X = 14.696 * 2 = 29.3938 cm.

Question 13

Find the ratio of the surfaces of the inscribed and circumscribed spheres about a cube.

Accepted Answer

4 × 3 . 14 × r 2 : 4 × 3 . 14 × r 2 = 1 : 2

Question 14

The volume of a rectangular solid is to be increased by 50% without altering its base. To what extent the height of the solid must be changed.

Accepted Answer

Let l, b the sides of the base, so that the area of the base = lb remains constant. Let ‘h’ be the height of the solid. Volume of the solid, V = l× b× h. Since the volume is to be increased by 50%

Question 15

A store owner is packing books into larger boxes that measure 25 by 42 by 60 inches. If the measurement of each book is 7 by 6 by 5 inches, then how many books can be placed in the box ?

Accepted Answer

Number of books that can be placed in the larger box is given by Volume of Box / Volume of book. Now, n = 25 x 42 x 60 / 7 x 6 x 5 = 63000 / 210 = 300. Therefore, 300 books can be in that box.

Question 16

How many squares with sides 1/2 inch long are needed to cover a rectangle that is 4 ft long and 4 ft wide ?

Accepted Answer

It takes four of those little squares -- each one 1/2 inch on a side -- to cover one square inch. That's because each one is 1/2 x 1/2, or 1/4 of a square inch. In a rectangle that is 48 inches by 48 inches, there are 2304 square inches.(1 ft = 12 inches) Since it takes 4 little squares to cover 1 sq inch, then it would need 2304 x 4 squares (1/2 inch on a side each) to cover a rectangle 4 feet by 4 feet. That is 9,216 of those squares.

Question 17

88 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be ?

Accepted Answer

Vol of silver = πR 2 h = 66 C.C Given D= 1mm => R = D/2 = 1/20 cm => 22 7 × 1 20 2 × h = 66 = 66 x 400 x (7/22) = 8400 cm = 84 mtrs.

Question 18

A rectangular block has the dimensions 5x6x7 cm it is dropped into a cylindrical vessel of radius 6cm and height 10 cm. If the level of the fluid in the cylinder rises by 4 cm, What portion of the block is immersed in the fluid ?

Accepted Answer

Since level of water increased in cylinder by height 4. This is because of the rectangular block . Therefore , area of rectangular block immersed in water is = 22 7 × 6 × 6 × 4 Thats why portion of block immersed in water is = (22/7 * 36 * 4) / total vol. of rectangle = (22/7 * 36 * 4)/ (7*5*6) = (22/7 * 24)/ 35.

Question 19

66 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be:

Accepted Answer

Let the length of the wire be h. Radius r =1/2 mm = 1/20 cm πr 2 h = 66 => h= (66 x 20 x 20) x (7/22) = 8400cm = 84m

Question 20

What is the total surface area of a right circular cone of height 14 cm and base radius 7 cm?

Accepted Answer

h = 14 cm, r = 7 cm. So, l = 7 2 + 14 2 = 245 = 7 5 c m Total surface area = πrl + πr 2 = 22 7 × 7 × 7 5 + 22 7 × 7 × 7 = 154 5 + 1 = (3.236 x 154) sq.cm = 498.35 sq.cm

Volume and Surface Area Questions